无法将C程序与strsep（）和getenv（）一起使用

Question

I had this working before, but I was using pointers. getenv() keeps crashing so I copied the results using sprintf(). Now I want to deliminate with : and print only the first occurrence. Please help!

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void main(void) {

    char buf[999];
    const char *token;

    // HTTP_PROXY == 8.8.8.8:8888, end result should print 8.8.8.8

    sprintf(buf, "%s", getenv("HTTP_PROXY"));
    *token = strsep(&buf, ":");
    printf("New result: %s\n", token);

}

Answer 1

Since strsep wants a pointer to pointer, you must pass a pointer to pointer, not a pointer to array. This is not the same thing; make a pointer, and assign it buf . Pass a pointer to that new pointer to strsep to fix the first problem.

The second problem is that since strsep returns a pointer, you need to assign it to token , not to *token :

char buf[999];
const char *token;
// HTTP_PROXY == 8.8.8.8:8888, end result should print 8.8.8.8
sprintf(buf, "%s", getenv("HTTP_PROXY"));
char *ptr = buf;           // Since ptr, is a pointer...
token = strsep(&ptr, ":"); // ...you can pass a pointer to pointer
printf("New result: %s\n", token);

Answer 2

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void) {
  char const *http_proxy = getenv("HTTP_PROXY");
  if (http_proxy == NULL) {
    fprintf(stderr, "HTTP_PROXY not set: default 8.8.8.8:8888");
    http_proxy = "8.8.8.8:8888";
  }

  char *cpy = strdup(http_proxy);
  char *token = strtok(cpy, ":");
  if (token == NULL) {
    fprintf(stderr, "wrong format");
    return 1;
  }
  do {
    printf("Token: %s\n", token);
  } while ((token = strtok(NULL, ":")) != NULL);
  free(cpy);
}